Precalculus (6th Edition) Blitzer

Consider the equation $y=\left| x+1 \right|$. Substitute $x=-4$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| -4+1 \right| \\ & =\left| -3 \right| \\ & =3 \end{align} Substitute $x=-3$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| -3+1 \right| \\ & =\left| -2 \right| \\ & =2 \end{align} Substitute $x=-2$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| -2+1 \right| \\ & =\left| -1 \right| \\ & =1 \end{align} Substitute $x=-1$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| -1+1 \right| \\ & =\left| 0 \right| \\ & =0 \end{align} Substitute $x=0$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| 0+1 \right| \\ & =\left| 1 \right| \\ & =1 \end{align} Substitute $x=1$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| 1+1 \right| \\ & =\left| 2 \right| \\ & =2 \end{align} Substitute $x=2$ and solve: \begin{align} & y=\left| x+1 \right| \\ & =\left| 2+1 \right| \\ & =\left| 3 \right| \\ & =3 \end{align}